chapter 3 molecules, compounds, and chemical...
TRANSCRIPT
Chapter 3
Molecules, Compounds, and
Chemical Composition
Elements and Compounds
Elements combine together to make an almost limitless number of compounds.
The properties of the compound are totally different
from the constituent elements.
Chemical Bonds
Compounds are made of atoms held together by bonds.
Chemical bonds are forces of attraction between atoms.
The bonding attraction comes from attractions between protons and electrons.
Bond TypesTwo general types of bonding between atoms found in
compounds, ionic and covalent.
Ionic bonds result when electrons have been transferred between atoms, resulting in oppositely charged ions that attract each other.
Generally found when metal atoms bond to nonmetal atoms
Covalent bonds result when two atoms share some of their electrons.
Generally found when nonmetal atoms bond together
Formation of an Ionic Compound
}Na0
Cl0
⤸Neutral Atoms Undergo
Electron Transfer
Na+Cl-
Charged Ions
An Orderly Aggregate Called an Ionic Crystal
Formation of an Covalent Compound
Formation of an Covalent Compound
Representative Covalent Compounds
Chemical Formulas Describe Compounds
A compound is a distinct substance that is composed of atoms of two or more elements.
We describe the compound by describing the number and type of each atom in the simplest unit of the compound.
Each element is represented by its letter symbol.
The number of atoms of each element is written to the right of the element as a subscript.
Polyatomic ions are placed in parentheses.(if more than one is present)
Types of Formula: Empirical FormulaAn empirical formula gives the relative number of atoms
of each element in a compound.
It does not describe how many atoms, the order of attachment, or the shape.
For example:
1) The empirical formula for the ionic compound fluorspar is CaCl2. This means that there is 1 Ca2+ ion for every 2 Cl− ions in the compound.
2) The empirical formula for the molecular compound oxalic acid is CHO2. This means that there is 1 C atom and 1 H atom for every 2 O atoms in the molecule.
Types of Formula: Molecular Formula
A molecular formula gives the actual number of atoms of each element in a molecule of a compound.
It does not describe the order of attachment, or the shape.
The empirical formula for the molecular compound oxalic acid is CHO2.
The actual molecular formula is C2H2O4
Types of Formula: Structural Formula
A structural formula uses lines to represent covalent bonds and shows how atoms in a molecule are connected or bonded to each other.
Structural Formulas of Oxalic Acid
O C
O
C
O
O HH
Molecular Models
Practice — Find the empirical formula for each of the following
The ionic compound that has two aluminum ions for every three oxide ions
arabinose, C5H10O5
pyrimidine
ethylene glycol
C
C
N
C
N
C
H
HH
H
Al2O3
CH2O
C2H2N
CH3O
Classifying Elements & Compounds
Atomic elementselements whose particles are single atoms
Molecular elementselements whose particles are multi-atom molecules
Molecular compoundscompounds whose particles are molecules
Ionic compoundscompounds whose particles are cations and anions
Molecular View of Elements and Compounds
ElementsMOST ELEMENTS
Single atoms are the constituent particles.The atoms may be physically attracted to each other, but
are not chemically bonded together.
A FEW ELEMENTS Molecules are the constituent particles.
The molecules are made of two or more atoms chemically bonded together by covalent bonds.
Molecular Elements
H2
Cl2
Br2
I2
7
7A
N2 O
2 F
2
20
Compounds
SOME COMPOUNDS Composed of ions arranged in a 3-dimensional pattern
These are called ionic compounds.
OTHER COMPOUNDS Composed of individual molecule units
Each molecule contains atoms of different elements chemically attached by covalent bonds
These are called molecular compounds.
Ionic vs. Molecular Compounds
Propane – contains individual C3H8
moleculesTable salt – containsan array of Na+ ions
and Cl- ions
Classify Each of the Following as Either an Atomic Element, Molecular Element, Molecular Compound,
or Ionic Compound
Aluminum, AlAluminum chloride, AlCl3
Chlorine, Cl2
Acetone, C3H6OCarbon monoxide, COCobalt, Co
atomic elementionic compound
molecular elementmolecular compound
molecular compoundatomic element
Formula Units vs. Molecules
Compound must have no total charge, therefore we must balance the numbers of cations and anions in a compound to get 0 charge.
Practice — What are the formulas for compounds made from the following ions?
Potassium ion with a nitride ion
Calcium ion with a bromide ion
Aluminum ion with a sulfide ion
K3N
CaBr2
Al2S3
K+ with N3−
Ca2+ with Br−
Al3+ with S2−
calcium bromide
aluminum sulfide
potassium nitride
Formula MassThe mass of an individual molecule or formula unit
Also known as molecular mass or molecular weight
Sum of the masses of the atoms in a single molecule or formula unit
mass of 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
mass of 1 formula unit of MgCl2 = 2(35.45 amu Cl) + 24.30 amu Mg = 95.20 amu
Molar Mass
The relative masses of molecules can be calculated from atomic masses.
Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu
1 mole of H2O contains 2 moles of H and 1 mole of O.
molar mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g
so the Molar Mass of H2O is 18.02 g/mole
Molar Mass of Compounds
Practice — How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00
g PbO2 mol PbO2
Pb = 1 x 207.2 = 207.2O = 2 x 16.00 = 32.00PbO2 = 239.2 g/mol
Example: Find the number of CO2 molecules in 10.8 g of dry ice
g CO2 mol CO2 molec CO2
1 mol CO2 = 44.01 g, 1 mol = 6.022 x 1023
Practice — How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2)
g PbO2 mol PbO2 units PbO2
1 mol PbO2 = 239.2 g,1 mol = 6.022 x 1023
Practice — What is the mass of 4.78 x 1024 NO2 molecules?
molecules mol NO2 g NO2
1 mol NO2 = 46.01 g, 1 mol = 6.022 x 1023
Percent Composition
Percent Composition
Percentage of each element in a compound by mass
Can be determined from
1. the formula of the compound2. the experimental mass analysis of the compound
Find the mass percent of Cl in C2Cl4F2
Practice — Determine the mass percent composition of the following CaCl2
Mass Percent as a Conversion Factor
If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na.
g Na g NaCl
Find the mass of sodium in 6.2 g of NaCl
g NaCl mol NaCl mol Na g Na
58.44 g NaCl
Empirical Formulas
Empirical Formula
Simplest, whole-number ratio of the atoms of elements in a compound
Can be determined from elemental analysis
Finding an Empirical Formula from % Composition
1. Convert the percentages to grams
2. Convert grams to moles
3. Write a pseudoformula using moles as subscripts
4. Divide all by smallest number of moles
5. Multiply all mole ratios by number to make all whole numbers
Example:Find the empirical formula of aspirin with the given mass percent composition
Given:" C = 60.00%" " " H = 4.48% " " " O = 35.53%
Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O"
g C g H g O
mol C mol H mol O
pseudoformulaCxHyOz
empirical formulaCxHyOz
Manipulate subscripts to obtain whole-number ratio
Calculate the moles of each element
Write a pseudoformula
C4.996H4.44O2.220
Find the mole ratio
C2.25H2.00O1.00Multiply subscripts by factor to give whole number
C9H8O4
(x 4)
÷ 2.220
C4.996H4.44O2.220
Practice — Determine the empirical formula of stannous fluoride, which
contains 75.7% Sn (118.70 g/mol) and the rest fluorine (19.00 g/mol)
g Sn mol Sn
g F mol F
pseudo-
formula
empirical
formula
mole
ratio
wholenumber
ratio
Given: 75.7% Sn, (100 – 75.3) = 24.3% F
ElementRatio in Grams
Molar MassRatio in Moles
Ratio in Moles
Sn 75.7g1mol
118.7g0.6377 1.000
F 24.3g1mol
19.00g1.279 2.005
Practice — Determine the empirical formula of stannous fluoride, which contains 75.7% Sn
(118.70 g/mol) and the rest fluorine (19.00 g/mol)
SnF2
X
X
Practice — Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85)
and the rest oxygen (16.00)
Given: 72.4% Fe, (100 – 72.4) = 27.6% O
g Fe mol Fe
g O mol O
pseudo-
formula
empirical
formula
mole
ratio
whole
number
ratio
Practice — Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)
ElementRatio in Grams
Molar MassRatio in Moles
Ratio in Moles
Ratio in Moles
Fe 72.4g1mol
55.85g1.296 1.000 3
O 27.6g1mol
16.00g1.725 1.33 4
Fe3O4
X
X
Molecular Formulas
Molecular Formulas
The molecular formula is a multiple of the empirical formula.
To determine the molecular formula you need to know theempirical formula and the molar mass of the compound.
Find the molecular formula of butanedione if its empirical formula is C2H3O and its molar mass (MM) is 86.03 g/mol.
Factor of 2
Practice – Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01)
Molecular formula = {C5H3} x 4 = C20H12
C5 =" 5(12.01 g) = 60.05 gH3 =" 3(1.01 g) = 3.03 gC5H3 "" = 63.08 g
252
?
Combustion Analysis
Combustion Analysis
A known mass of compound is burned in oxygen and the masses of the products formed (CO2 and H2O) are determined.
By knowing the masses of the products and composition of constituent elements in the product, the original amount of constituent elements can be determined.
It is assumed that all of the carbon in the original sample is converted to carbon dioxide and all of the hydrogen in the sample is converted to water.
(Generally used for organic compounds containing C, H, O)
Combustion Analysis
Example of Combustion AnalysisCombustion of a 0.8233 g sample of a compound containing only
carbon, hydrogen, and oxygen produced the following:
CO2 = 2.445 g H2O = 0.6003 g
! Determine the empirical formula of the compound.
g
CO2, H2O
mol
ratio
empirical
formula
mol
CO2, H2O
mol
C, H
g
C, H
g
O
mol
O
mol
C, H, O
pseudo
formula
This came from C.This came from H.
1 mole CO2 = 44.01 g CO2 1 mole H2O = 18.02 g H2O
1 mole C = 12.01 g C1 mole H = 1.008 g H 1 mole O = 16.00 g O
1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H
In the original sample
In the original sample
In the original sample
C0.05556H0.06662O0.00556
Pseudo formula
÷ 0.00556
Empiricalformula
Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2.
If the molar mass of caproic acid is 116.2 g/mol,what is the molecular formula of caproic acid?
moles
g
0.0145 0.0870 0.0436
0.232 0.0877 0.524
O H C
Molecular formula = {C3H6O} x 2 = C6H12O2